Results 1 to 4 of 4

Math Help - Proving identities

  1. #1
    Newbie
    Joined
    Jan 2007
    Posts
    6

    Proving identities

    I have absolutely no idea how this works, so if you can go through these two for me maybe I can do the other 8

    A) x(2) + 2x -2 = (x+1)(x-1) +2(x+1)-3

    b) 6x + 9 = 7(x+1) (x-2)

    thank you for any help
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    10,212
    Thanks
    419
    Awards
    1
    Quote Originally Posted by Kim2425 View Post
    I have absolutely no idea how this works, so if you can go through these two for me maybe I can do the other 8

    A) x(2) + 2x -2 = (x+1)(x-1) +2(x+1)-3

    b) 6x + 9 = 7(x+1) (x-2)

    thank you for any help
    The first IS an identity, the second isn't. Are you supposed to test if the equation is an identity?

    a) There is nothing to do on the LHS, so go to the RHS and simplify:
    (x + 1)(x - 1) + 2(x + 1) - 3

    x^2 - 1 + 2x + 2 - 3

    x^2 + 2x + (-1 + 2 - 3)

    x^2 + 2x - 2

    Since the LHS and the RHS are identical, this equation is an identity.

    b) Again, nothing to do on the LHS, so go to the RHS:
    7(x + 1)(x - 2)

    7(x^2 - x - 2)

    7x^2 - 7x - 14

    The LHS and RHS are not the same, so this is not an identity, but a "conditional equation."

    -Dan
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Aug 2006
    Posts
    56
    6x + 9 = 7(x+1) (x-2)
    Expanding the right side, and rewriting yields:

    6x + 9 = 7x^2 - 7x -14

    How can this possibly be an identity? Perhaps you were given a list of equations and you are supposed to pick out the ones that are identities from the ones that are not?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,914
    Thanks
    779
    Hello, Kim!

    You're supposed to show that the two sides are equal.
    . . How hard can that be?
    And you have a typo in the second one . . .


    A) .x + 2x - 2 .= .(x + 1)(x - 1) +2(x + 1) - 3
    Simplify the right side: .x - x + x - 1 + 2x + 2 - 3 .= .x +2x - 2

    . . and it equals the let side . . . get it?



    B) . 6x + 9 .= .7(x + 1) - (x - 2)
    Now you try this one . . .

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: June 23rd 2010, 12:59 AM
  2. help with proving a few identities
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: May 1st 2010, 12:32 AM
  3. Proving identities
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: April 4th 2010, 09:19 PM
  4. proving identities....
    Posted in the Trigonometry Forum
    Replies: 1
    Last Post: December 2nd 2009, 11:21 PM
  5. Proving identities
    Posted in the Trigonometry Forum
    Replies: 4
    Last Post: December 2nd 2009, 08:08 AM

Search Tags


/mathhelpforum @mathhelpforum